leaching - lecture notes on unit operations equipment shown in figure 1 constitutes one stage of...
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Leaching
All rights reserved. Armando B. Corripio, PhD, PE. 2014
Contents Leaching....................................................... 1
1 The Leaching Operation ..................................... 2
2 Balances on Leaching Battery ............................... 3
2.1 Underflow Function. .................................... 5
2.2 Specifications. ........................................ 5
3 Operating Line ............................................. 6
4 Equilibrium Line ........................................... 7
5 Flow Basis ................................................. 7
6 Balances on the First Stage ................................ 9
Summary ..................................................... 19
Review Questions ............................................ 19
Problems .................................................... 20
Leaching is the extraction of a solute from an insoluble solid
by contact with a liquid solvent. The following are examples of
leaching:
Extraction of sugar from sugar cane by contact with hot
water
Extraction of mineral salts from crushed ores
Extraction of oil from halibut livers by contact with
benzene
Extraction of sodium hydroxide from insoluble carbon
carbonate by contact with water
Extraction of coffee from coffee grounds by contact with
hot water.
1 The Leaching Operation
Leaching is carried out by contacting the solid with the liquid
solvent and separating the liquid extract from the solid.
Various types of equipment can be used to carry out this
operation, but the most common is to do the contacting in a
digester followed by separating the extract in a decanter, as
Figure 1 shows.
Solvent
Extract Out
Solid In
Solid Out
Digester
Decanter
Figure 1. Typical leaching equipment.
The equipment shown in Figure 1 constitutes one stage of
leaching. The complete operation usually requires a battery
consisting of a number of stages arranged in countercurrent
flow, as shown in Figure 2.
Stage1
Stage2
Stage3
Stage4
Stage5
Solvent In Extract Out
Solid Out Solid In
Figure 2. Cascade of countercurrent stages for leaching.
2 Balances on Leaching Battery
A leaching operation involves three different components, the
solid, the solute, and the solvent. The solid is insoluble in
the solvent so its flow does not change through the battery. It
is the same from stage to stage and the same from inlet to
outlet. The solid, however carries with it some of the solution
(solute + solvent) from stage to stage. The flow of the solution
carried by the solid is called the underflow. The extract that
flows from stage to stage is called the overflow and consists of
solution only, solute plus solvent. To avoid mistakes in the
leaching material balances it is very important to understand
the following:
The UNDERFLOW is the solution, solute plus solvent,
carried by the solid from stage to stage. IT DOES NOT
INCLUDE THE SOLID.
As in any other operation the design calculations start
with the over-all material balances around the entire battery.
Consider the schematic of the battery shown in Figure 3:
Cascade
Va
ya
La
xa
Vb
yb
Lb
xbFeed
SolventExtract
Figure 3. Over-all balances on the cascade
The notation of Figure 3 is as follows:
L = flow of underflow, solvent plus solute
V = flow of overflow
x = composition of solute in the underflow
y = composition of solute in the overflow
Subscript a is normally used on the end of the cascade where the
solid enters and the extract exits, and subscript b for the end
where the solid exits and the solvent enters. The compositions x
and y are of the solute in the solutions, excluding the solid.
As the solid flow through the cascade is constant, there is
no need to write a balance on the solid, so this leaves us two
components, the solute and the solvent, and two balances.
Usually we write a total balance and a solute balance, but
remember that the total balance does not include the solid.
Total balance: La + Vb = Lb + Va (1)
Solute balance: Laxa + Vbyb = Lbxb + Vaya (2)
2.1 Underflow Function.
One additional relationship in leaching is the dependence of the
underflow L on its composition x. The underflow is the liquid
solution entrained in the solids leaving the decanter (see
Figure 1). In general the underflow is a function of its
composition as the composition affects the density and viscosity
of the solution, that is, the underflow may be larger the higher
its composition. The underflow function, U(x), must be
determined experimentally and may be a function of the
properties of the solid, of the solution, and of the equipment
used to separate the overflow (the decanter). As will be
discussed shortly, under certain conditions and on the
appropriate flow basis the underflow may be assumed to be
constant, that is, to remain the same from stage to stage.
2.2 Specifications.
With two balance equations plus the underflow function and eight
variables, solution of the over-all balances requires five
specifications. The feed flow and composition, La and xa, and the
solvent composition, yb, are standard specifications in a design
problem. Another specification is either the solute recovery or
the exit underflow composition, xb; finally we have either the
solvent inlet rate Vb or the exit extract composition, ya, but
one of these is usually a design variable left to the discretion
of the designer.
After the compositions and flows of the outlet streams are
determined from the over-all balances, the number of required
stages can be determined from stage to stage calculations using
the operating and equilibrium lines. These will be discussed
next.
3 Operating Line
As usual, the operating line is obtained from balances around n
stages in the cascade, as follows:
Stage1
Stage2
Stagen
StageN
Va
ya
Vb
yb
La
xa
V2
y2
V3
y3
Vn+1
yn+1
VN
yN
L1
x1
L2
x2
Ln-1
xn-1
Ln
xn
LN-1
xN-1
Lb
xb
Vn
yn
Figure 4. Balances around n stages to obtain the operating line.
Once again we can write a total balance and a solute balance:
Total balance: La + Vn+1 = Va + Ln (3)
Solute balance: Laxa + Vn+1yn+1 = Vaya + Lnxn (4)
Combine the two equations and solve for yn+1, to obtain the
equation of the operating line:
(5)
The operating line goes through the points (xa,ya) and (xb,yb).
When the underflow L can be assumed constant, by Equation (3),
the overflow V is also constant and the operating line is
straight. When the underflow varies from stage to stage the
operating line is curved and Equation (5) must be used to plot
intermediate points on the line.
4 Equilibrium Line
In leaching the equilibrium assumption is that the solution
entrained in the solid leaving a stage has the same composition
as that of the extract leaving the stage, that is,
yn = xn (6)
Thus the equilibrium line is always a straight line with a slope
of 1.0. Notice that this is valid independent of the basis used
on the composition (mole fraction, weight fraction, molar ratio,
weight ratio, or concentration).
5 Flow Basis
As with other stagewise operations, the determination of the
number of required stages is simplified when both the
equilibrium and operating lines are straight, because the number
of stages can then be determined with the Kremser equation.
Given that, in leaching, the equilibrium line is straight for
any flow basis, it behooves a smart chemical engineer to select
a flow basis in which the operating line is straight, if
possible. For example, if the volume of solution entrained in
the solid is approximately constant, then a volumetric flow
basis (liters/min) results in a constant underflow and thus a
straight operating line.
It is important to realize that the flow basis determines
the basis for the composition. If the basis for the composition
does not match the basis for the flow, the calculation of the
terms Fx in the balance equations will not result in the rate of
solute as it must. For example, F in liters/min must be
multiplied by x in kg/liter (concentration) to obtain kg/min of
solute.
The following are some of the common flow and composition
bases that can be selected to carry out the material balances.
It is important to realize that the flow basis and the
composition basis must match:
Flow basis Composition basis
Weight flow Weight fraction
Molar flow Mole fraction
Solute-free weight flow Weight ratio
Solute-free molar flow Molar ratio
Volume flow Concentration_____
On the solute-free basis the flow is in rate of solvent per unit
time and the composition is in units of solute per unit of
solvent, for example, kg of solvent/hr and kg solute/kg solvent.
The conversions of total flow to solute-free basis and weight
fraction to weight ratio are as follows:
L’ = L(1 – x)
V’ = V(1 – y)
Where L’ and V’ are the flows in weight of solvent per unit
time, and X and Y are the weight ratios in weight of
solute/weight of solvent.
6 Balances on the First Stage
The feed to the battery does not follow the underflow
relationship because it is not the exit stream from a decanter
as those in the battery. Therefore the feed point is not on the
operating line determined for the rest of the stages making it
necessary to carry out separate balance calculations for the
first stage, as follows:
Figure 5. Balances on the First Stage.
The first stage is assumed to be an equilibrium stage, as are
the others:
x1 = ya
The underflow from stage 1 follows the underflow relationship:
L1(x1). Then
V2 = Va + L1 - La
Va
ya
Vb
yb
La
xa
Lb
xb
V2
y2
L1
x1
1 Stages 2 to N
The determination of stages 2 to N can now be carried out by
stage-to-stage calculations, McCabe-Thiele graphical procedure
or, if the equilibrium and operating lines are straight, the
Kremser equation.
Example 1. Design of a Leaching Battery
The feed to a leaching battery consists of 10,000 kg/hr
containing 70 weight% insoluble solid and the balance solute.
Pure solvent at the rate of 9,000 kg/hr is used to extract 95%
of the solute. Each kg of solid carries (1.1 + 0.25x) kg of
solution of solute mass fraction x. Determine the number of
equilibrium stages required to carry out the desired extraction.
Solution: The first thing to do is to draw a schematic of the
battery showing all the problem data.
Cascade
Va
ya
La
xa
Vb
yb
Lb
xbFeed
SolventExtract
10,000 kg/hr70 wt% solids30 wt% solute
= 9,000 kg/hr
= 0 (pure)
95% recovery
U(x) = 1.1 + 0.25x
Figure 6. Schematic and problem data for Example 1.
The feed consists of the following:
Solids: 0.70(10,000 kg/hr) = 7,000 kg/hr
Solute: 0.30(10,000 kg/hr) = 3,000 kg/hr
Solvent: 10,000 – 7,000 – 3,000 = 0 kg/hr
Total solution in feed: La = 3.000 + 0 = 3,000 kg/hr
Composition of the solution in the feed: xa = 3,000/3,000 = 1.0
So the composition of the solute in the solution in the
feed is 100 weight% (not 30 weight%). This is because the solids
are excluded and there is no solvent in the feed. (You must
study this until you are sure you understand it.)
The underflow function given in the statement of the
problem is in kg of solid and is normally obtained
experimentally from measurements on the equipment. So the flow
of underflow solution leaving each stage is calculated by:
Ln = (7,000 kg/hr)(1.1 + 0.25xn)
From the recovery we can determine the flow of unrecovered
solute leaving with the solids:
Lbxb = (1 – 0.95)Laxa = 0.05(3,000)(1.0) = 150 kg/hr
Because the underflow from the last stage follows the underflow
function, Lb = 7,000(1.1 + 0.25xb), and we can write:
Lbxb = 7,000(1.1 +0.25xb)xb = 150 kg/hr
We can solve this equation to obtain: xb = 0.0194
Then, Lb = 7,000(1.1 + 0.25×0.0194) = 7,730 kg/hr
From the total balance: Va = La + Vb – Lb
= 3,000 + 9,000 – 7,730 = 4,270 kg/hr
And from the solute balance:
( )( ) ( )( ) ( )( )
We must now carry out the balances around the first stage:
Stages 2 to N
Va = 4270 kg/hrya = 0.667
La = 3,000 kg/hrxa = 1.0
Vb = 9,000 kg/hryb = 0
Lb = 7,730 kg/hrxb = 0.0194
L1
x1 = ya = 0.667
V2
y2
1
Figure 7. Balances on stage 1 for Example 1.
Assuming equilibrium stages, x1 = ya = 0.667.
From underflow function: L1 = 7,000(1.1 + 0.25x1)
= 7,000(1.1 + 0.25×0.667)
= 8,870 kg/hr
Total balance on stage 1: V2 = Va + L1 – La
= 4,270 + 8,870 -3,000
= 10,140 kg/hr
Solute balance on stage 1:
( ) ( ) ( )
Now we are ready to determine stages 2 to N. First let us check
if the operating line can be assumed to be approximately
straight. We do this by comparing the slope of the operating
line at both ends:
L1/V2 = 8,870/10,140 = 0.875 Lb/Vb = 7,730/9,000 = 0.858
As the slope changes by less than 2% from one end to the other,
we can assume the operating line is approximately straight. As
the equilibrium line is also straight, we can use the Kremser
equation to estimate the required number of equilibrium stages
after the first one. For leaching:
x1* = y2 = 0.568 xb
* = yb = 0
(
)
(
)
(
)
(
)
Notice that the Kremser equation is applied for stages 2 to N,
that is, it does not include the first stage. This is the reason
the “1” is added.
Had we determined the required number of equilibrium stages
by the McCabe-Thiele method, we would have obtained the
following plot:
Figure 8. McCabe-Thiele Diagram for Example 1.
As expected, the number of equilibrium stages is a bit over 13,
showing that the Kremser equation is a good approximation in
this example. The graph shows why the first stage must be
considered separately.
Example 2. Design of a Sugar Cane Mill
The mills in a sugar Factory are fed 30,000 kg/hr of cane
containing 10 weight% fiber, 13 weight% sugar, and the balance
water. Pure hot water is used to extract the sugar. The fiber,
called bagasse, is insoluble in the water. It leaves each stage
carrying 1.2 kg of sugar solution per kg of fiber on a sugar-
free basis. Design the battery to recover 99.5% of the sugar in
the cane and to produce a final raw juice containing 12 weight%
sugar.
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
1.000
0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000
x, kg oil/kg solution
y, kg
oil/k
g s
olu
tio
n
Equilibrium
Operating
Stages
Figure 9. Rendition of a sugar mill with three mills.
Solution: The first step, as always, is to draw a schematic of
the mill showing the problem data.
Cascade
Va
ya = 0.12
La
xa
Vb
yb
Lb
xbFeed
SolventExtract
30,000 kg/hr10 wt% fibers (solids)13 wt% sugar (solute)
Balance water (solvent)
= 0 (pure)
99.5% recovery
U(x) = 1.2 (sugar-free basis)
Figure 10. Schematic of Sugar Mill of Example 2.
The feed consists of:
Stage 1
RawJuice
Bagasse
Solvent(Hot water)
Cane
Stage 2 Stage 3
Fiber (insoluble solid): 0.10(30,000 kg/hr) = 3,000 kg/hr
Sugar (solute): 0.13(30,000 kg/hr) = 3,900 kg/hr
Water (solvent): 30,000 – 3,000 – 3,900 = 23,100 kg/hr
By the statement of the problem, the underflow is constant
on a sugar-free (solute-free) basis, so the problem is easier to
solve on this basis. The basis consists of measuring all flows
on sugar-free basis, which is the water weight flow, and the
compositions in weight ratios, or kg of sugar per kg of water.
As follows:
Feed: La’ = 23,100 kg/hr Xa = 3,900/23,100 = 0.169
Extract: Ya = 0.12/(1 – 0.12) = 0.136 kg sugar/kg water
Underflow: L’ = 3,000U(X) = 3,000(1.2) = 3,600 kg/hr (constant)
From the recovery, the unrecovered sugar leaving in the bagasse
is:
Lb’Xb = (1 – 0.995)La’Xa = 0.005(23,100)(0.169) = 19.5 kg/hr
Now, Lb’ = L’ = 3,600 kg/hr, and Xb = 19.5/3,600 = 0.00542
From the solute balance:
( )
From the total balance: Vb’ = Va’ + Lb’ – La’
= 28,600 + 3,600 – 23,100
= 9,100 kg/hr
Because the underflow is constant on a sugar-free basis, the
overflow is also constant on a sugar-free basis, that is,
V’ = Vb’ = 9,100 kg/hr
Now that we have all the outlet flows and compositions and the
solvent rate, we are ready to determine the required number of
equilibrium stages. We start by doing the balances on stage 1:
Stages 2 to N
Va’ = 28,600 kg/hrYa = 0.136
La’ = 23,100 kg/hrXa = 0.169
Vb’ = 9,100 kg/hrYb = 0
Lb’ = 3,600 kg/hrXb = 0.00542
L’ = 3,600 kg/hrX1 = Ya = 0.136
V’ = 9,100 kg/hrY2
1
Figure 12. Balances on Stage 1 for Example 2.
As stage 1 is an equilibrium stage, as are the others,
X1 = Ya = 0.136
Solute balance on stage 1:
( ) ( ) ( )
As the underflow and overflow rates from stages 2 to N are
constant on a sugar-free basis, the operating line is straight
on this basis. In leaching the equilibrium line is straight on
any basis (Y = X), so we can use the Kremser equation to
determine the number of required equilibrium stages from stages
2 to N:
X1* = Y2 = 0.0522 Xb* = Yb = 0
(
)
(
)
(
)
(
)
Thus four equilibrium stages are required.
The weight fractions of the outlet streams can be
calculated from the weight ratios:
xb = Xb/(1 + Xb) = 0.00542/1.00542 = 0.00539
ya = Ya/(1 + Ya) = 0.136/1.136 = 0.120
The outlet flows in total solution weight rates are:
Lb = Lb’(1 + Xb) = 3,600(1.00542) = 3,620 kg/hr
Va = Va’(1 + Ya) = 28,600(1.136) = 32,500 kg/hr
It is very important to understand the following points:
The balance calculations for this example can be carried
out on a total solution flow basis and the corresponding
compositions in weight fractions. The results will be
exactly the same as the ones obtained here.
Given the conditions from the problem statement, the
operating line is not straight on the total solution flow
basis, so if the weight fractions are used in the Kremser
equation the resulting number of equilibrium stages will
only be approximate.
Exact results can be obtained by converting the weight
fractions to weight ratios and then using the weight ratios
in the Kremser equation.
If you are not fully convinced you should try it and see that
you get a different result when you use the weight fractions and
the same result when you use the weight ratios. This will be a
good exercise.
Summary
These notes have presented the design of leaching batteries. It
has been shown that what is characteristic of leaching is that
the equilibrium line is always straight with a slope of 1.0. As
this is true on any composition basis it is possible under
certain conditions to pick the basis that produces a straight
operating line thus simplifying the calculation of the required
number of equilibrium stages.
Review Questions
1. Define the leaching operation.
2. What is the underflow? Does the underflow include the
insoluble solids?
3. What is the underflow function? How it is normally obtained
in practice?
4. How many over-all mass balances are independent in
leaching?
5. How many specifications are required to solve the over-all
mass balance equations? Enumerate them.
6. What is the equilibrium relationship in leaching? On what
composition basis is it valid?
7. How is the basis for the stream flows and composition
selected?
8. Why must separate balances be made on the first stage of a
leaching battery?
Problems
1. Extraction of Oil from Meal. Oil is to be extracted from
meal, an insoluble solid, using benzene as the solvent. The
unit is to treat 1,800 kg/hr of meal consisting of 70
weight% meal and 30 weight% oil. The solvent enters at 855
kg/hr and contains 1.5 weight% oil and the balance benzene.
It is desired to recover 85% of the oil in the feed.
Experiments carried out with the battery have resulted in
the relationship below between the solution retained by the
solid (underflow) and the oil concentration of the
solution.
U(x) = 0.49842 + 0.05845x + 0.16786x2
where U(x)is the kg of underflow per kg of solid and x is
the weight fraction oil in the underflow. Draw a schematic
of the battery showing all the problem data and design the
battery to determine the required number of equilibrium
stages. Report also the flows and compositions of all the
streams in and out of the battery. State all assumptions.
2. Extraction of Coffee in a Countercurrent Leaching Battery.
A countercurrent leaching battery is to extract coffee from
coffee grounds using hot water. The feed, at 1,100 kg/hr,
contains 42 weight% coffee and the balance insoluble solid
(coffee grounds). It is desired to recover 99.5% of the
coffee in the feed. The grounds retain 0.45 kg of solution
per kg of insoluble solid and the extract leaving the
battery is to contain 55 weight% coffee. Draw a schematic
of the battery showing all the problem data and design the
battery to determine the number of required equilibrium
stages and the flows and compositions of all the streams in
and out of the battery.
3. Leaching of NaOH from Calcium Carbonate Sludge. (Based on
Problem 23.2, McCabe, Smith & Harriott, 7th ed., page 792.)
It is desired to extract the sodium hydroxide from the
sludge produced by the following reaction:
Na2CO3 + CaO + H2O --> CaCO3 + 2NaOH
The insoluble CaCO3 carries with it 1.5 kg of solution per
kg of CaCO3 in flowing from one stage to the next. It is
desired to recover 99% of the NaOH. The reactor products
enter the battery with no excess reactants but with 0.6 kg
of water per kg of CaCO3. The solvent is pure water. Draw a
schematic of the battery showing all the problem data and
determine the composition of the extract from the battery
and the required number of equilibrium stages. Hint: Notice
that the sludge contains 1 kmole of CaCO3 (MW = 100) and 2
kmole of NaOH (MW = 40).
4. Dewaxing of Paper with Kerosene. A countercurrent leaching
battery is to be designed to extract wax from insoluble
paper pulp using kerosene as the solvent. The feed consists
of 90 kg of wax paper per hour containing 25 weight% wax
and the balance pulp. Solvent is fed containing 0.05
weight% wax and the balance kerosene. It is desired to
leave no more than 0.2 kg of wax per 100 kg of pulp in the
extracted solid. The underflow leaving each stage contains
2 kg of solution per kg of pulp on a solute-free basis.
Draw a schematic of the battery showing all the problem
data and design the battery to determine the number of
required equilibrium stages and the flows and compositions
of all the streams in and out of the battery.
5. Countercurrent Leaching Battery for Copper Ore. (Based on
McCabe, Smith & Harriott, 7th edition, Problem 23.1, page
792.) The feed to a countercurrent leaching battery is
8,500 kg/hr of roasted copper ore consisting of 21.5
weight% CuSO4 and the balance inert rock. It is desired to
extract 95% of the CuSO4 using pure water as a solvent.
Experimental data reveals that the underflow from each
stage contains a constant amount of 1.8 kg of solution per
kg of inert rock on a solute-free basis. If the solution
exiting the battery is to contain 15weight% CuSO4,
determine the required number of equilibrium stages. Draw a
schematic of the battery showing all the problem data and
report the flows and compositions of all the streams in and
out of the battery.
6. Parallel Leaching Battery for Copper Ore. (Based on McCabe,
Smith & Harriott, 7th edition, Problem 23.1, page 792)
Using the feed rate and composition of the countercurrent
battery of Problem 5 consider a parallel leaching battery
with the same number of stages and feeding the amount of
solvent in each stage as is fed to the countercurrent
battery. Determine the composition of the final solution
carried by the rock, the composition of the combined
extract from all the stages after they are mixed, and the
recovery of CuSO4. Assume the underflow relationship is the
same.